Water is leaking out of an inverted conical tank at a rate of 10,500 cm3/min at the same time that water is being pumped into the tank at a constant rate. The tank has height 6 m and the diameter at the top is 4 m. If the water level is rising at a rate of 20 cm/min when the height of the water is 2 m, find the rate at which water is being pumped into the tank
<h3><u>Answer:</u></h3>
The rate at which water is being pumped into the tank is 289,752 ![cm^3/min](https://tex.z-dn.net/?f=cm%5E3%2Fmin)
<h3><u>Solution:</u></h3>
According to question,
There is an inverted conical tank, through which water is leaking at a rate of 10,500 cm3/min at the same time that water is being pumped into the tank at a constant rate
<em><u>The dimension of tank are:
</u></em>
Diameter = 4cm
Radius(r) =
= 2cm
Height = 6cm
Clearly we can see that height is 3 times radius so, we can write
h = 3r OR r = h/3 ……………………. (1)
<em><u>The volume of cone "V" is given as:</u></em>
-------- (2)
From (1) and (2)
![\text { Volume of cone(V) }=\frac{1}{3} \pi\left(\frac{h}{3}\right)^{2} h](https://tex.z-dn.net/?f=%5Ctext%20%7B%20Volume%20of%20cone%28V%29%20%7D%3D%5Cfrac%7B1%7D%7B3%7D%20%5Cpi%5Cleft%28%5Cfrac%7Bh%7D%7B3%7D%5Cright%29%5E%7B2%7D%20h)
-------- (3)
<em><u>Now we calculate the derivate:- </u></em>
![\frac{d V}{d t}=\frac{3 \pi h^{2}}{27} \frac{d h}{d t}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%20V%7D%7Bd%20t%7D%3D%5Cfrac%7B3%20%5Cpi%20h%5E%7B2%7D%7D%7B27%7D%20%5Cfrac%7Bd%20h%7D%7Bd%20t%7D)
--------- (4)
According to question, when height is 2m = 200cm, the water level is rising at a rate of 20 cm/min
![\frac{d h}{d t}=20 \mathrm{cm} / \mathrm{min}](https://tex.z-dn.net/?f=%5Cfrac%7Bd%20h%7D%7Bd%20t%7D%3D20%20%5Cmathrm%7Bcm%7D%20%2F%20%5Cmathrm%7Bmin%7D)
On putting above values in equation(4) and solving we get
![\frac{d V}{d t}=279252](https://tex.z-dn.net/?f=%5Cfrac%7Bd%20V%7D%7Bd%20t%7D%3D279252)
Hence, the rate at which water is being pumped is 289,752
which is the sum of water volume increasing at rate of 279,252 and 10,500 leaking out